Bài 1 trang 75 SGK Toán 11 Tập 2 Cánh diều - CD

a) Ta có:

\(\begin{array}{l}y' = \left( {\frac{1}{{2x - 3}}} \right)' = - \frac{2}{{{{\left( {2x - 3} \right)}^2}}}\\ \Rightarrow y'' = \left( { - \frac{2}{{{{\left( {2x - 3} \right)}^2}}}} \right)' = \left( { - 2.\frac{1}{{\left( {2x - 3} \right){}^{^2}}}} \right)' = - 2\frac{{\left( {{{\left( {2x - 3} \right)}^2}} \right)'}}{{{{\left( {2x - 3} \right)}^4}}} = - 2\frac{{2.\left( {2x - 3} \right).2}}{{{{\left( {2x - 3} \right)}^4}}}\\ = - \frac{8}{{{{\left( {2x - 3} \right)}^3}}}\end{array}\)

b) Ta có:

\(\begin{array}{l}y' = \left( {{{\log }_3}x} \right)' = \frac{1}{{x\ln 3}}\\ \Rightarrow y'' = \left( {\frac{1}{{x\ln 3}}} \right)' = - \frac{{\left( {x\ln 3} \right)'}}{{{{\left( {x\ln 3} \right)}^2}}} = - \frac{{\ln 3}}{{{{\left( {x\ln 3} \right)}^2}}} = - \frac{{\ln 3}}{{{{\left( {x\ln 3} \right)}^2}}} = - \frac{1}{{x.\ln 3}}\end{array}\)

c) Ta có:

\(\begin{array}{l}y' = \left( {{2^x}} \right)' = {2^x}.\ln 2\\ \Rightarrow y'' = \left( {{2^x}.\ln 2} \right)' = {2^x}.\ln 2.\ln 2 = {2^x}.{\left( {\ln 2} \right)^2}\end{array}\)

-- Mod Toán 11 HỌC247